"Fat" Cantor Set So the standard Cantor set has an outer measure equal to $0$, but how can you construct a "fat" Cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of $1$. I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of $0$ as well...
Are there other constraints that need to be made in order to accomplish this? 
 A: Say you delete the middle third.
Then delete two intervals the sum of whose lengths is $1/6$ from the two remaining intervals.
Then delete intervals the sum of whose lengths is $1/12$, one from each of the four remaining intervals.
And so on.  The amount you delete is $\displaystyle\frac13+\frac16+\frac{1}{12}+\cdots= \frac23.$  That is less than the whole measure of the interval $[0,1]$ from which you're deleting things.
A: The measure of a Cantor set cannot be 1, but certainly can be arbitrary close to it.  Let the total measure of the pieces removed at step $n$ be equal $m_n$, so we remove an open intervalof length $m_1$ in the first step, then remove two open intervals of total length  $m_2$ from the two pieces remaining after step 1.  Then remove four open intervals of total length $m_3$ from the remaining four pieces left after step 2.  Continuing this way, we remove open intervals with total measure $M=m_1+m_2+...$. The measure of the compact set $C$ left is $1-M$.  If $M=1$, the measure of $C$ is $0$, if $M<1$ then the measure of $C$ is positive---it is fat.
